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erma4kov [3.2K]
3 years ago
10

If the work done by a battery to move a unit charge through the battery from one terminal to the other is doubled, what is the e

ffect on the battery's emf?a. emf becomes zero.b. emf is halved.
Physics
1 answer:
gtnhenbr [62]3 years ago
6 0

The emf is doubled

Explanation:

The work done by a battery to move a unit charge through the battery from one terminal to the other is equal to the change in electric potential energy of the charge, therefore it can be written as

W=q\Delta V

where

q is the charge

\Delta V is the potential difference through which the charge has been moved

Since the charge is moving through the battery, this means that the potential difference is equal to the emf of the battery \epsilon, therefore

W=q\epsilon (1)

In this problem, we are told that the work done by the battery is doubled, so

W'=2W

From eq.(1), we see that the emf of the battery is proportional to the work done: therefore, since the work doubles, the battery's emf doubles as well.

Learn more about potential difference:

brainly.com/question/4438943

brainly.com/question/10597501

brainly.com/question/12246020

#LearnwithBrainly

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Answer:

a) h=250\ m

b) \Delta h=0.0835\ m

Explanation:

Given:

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a)

<u>Maximum height reached by the helicopter:</u>

using the equation of motion,

h=u.t+\frac{1}{2} a.t^2

where:

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t = time of observation

h=0+0.5\times 5\times 10^2

h=250\ m

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<u>height fallen freely by Austin:</u>

h_f=u.t_f+\frac{1}{2} g.t_f^2

where:

u= initial velocity of fall at the top = 0 (begins from the max height where the system is momentarily at rest)

t_f= time of free fall

h_f=0+0.5\times 9.8\times 7^2

h_f=240.1\ m

<u>Velocity just before opening the parachute:</u>

v_f=u+g.t_f

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v_f=68.6\ m.s^{-1}

<u>Time taken by the helicopter to fall:</u>

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where:

u= initial velocity of the helicopter just before it begins falling freely = 0

t_h= time taken by the helicopter to fall on ground

h= height from where it falls = 250 m

now,

250=0+0.5\times 9.8\times t_h^2

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From the above time 7 seconds are taken for free fall and the remaining time to fall with parachute.

<u>remaining time,</u>

t'=t_h-t_f

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<u>Now the height fallen in the remaining time using parachute:</u>

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h'=68.6\times 0.1428+0.5\times 2\times 0.1428^2

h'=9.8165\ m

<u>Now the height of Austin above the ground when the helicopter crashed on the ground:</u>

\Delta h=h-(h_f+h')

\Delta h=250-(240.1+9.8165)

\Delta h=0.0835\ m

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